Let $n$ be a fixed positive integer, and $$ \begin{cases} y_1+y_2+\ldots+y_n=x_1+x_2+\ldots+x_n,\\ y_1^2+y_2^2+\ldots+y_n^2=x_1^2+x_2^2+\ldots+x_n^2,\\ y_1^3+y_2^3+\ldots+y_n^3=x_1^3+x_2^3+\ldots+x_n^3,\\ \ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots,\\ y_1^n+y_2^n+\ldots+y_n^n=x_1^n+x_2^n+\ldots+x_n^n. \end{cases} $$
Is it true that then $(y_1,y_2,\ldots,y_n)$ is a permutation of $(x_1,x_2,\ldots,x_n)$? How to prove it?
Context: From comments:
--What did you try? How about $n=2,3$?
--Yes, I tried for $n=2$. It is true for $n=2$.
--For $n=3$: use $$x^3_1+x^3_2+x^3_3=(x_1+x_2+x_3)^3−3(x_1x_2+x_1x_3+x_2x_3)⋅(x_1+x_2+x_3)+ 3x_1x_2x_3,$$
$$x^2_1+x^2_2+x^2_3=(x_1+x_2+x_3)^2−2(x_1x_2+x_1x_2+x_2x_3),$$ same for $y$′s to show that $x_1,x_2,x_3$ and $y_1,y_2,y_3$ are roots of the same cubic equation.
The statement is true for every $n$. Suppose $x_1,...,x_n$ satisfy $x_1^j+...+x_n^j=a_j$ for $j=1,...,n$. We will find a monic polynomial of degree $n$, $f(x)=x^n-s_1x^{n-1}+s_2x^{n-2}-...$ whose roots are precisely $x_1,...,x_n$. We know that $s_i$ is the value of the $i$th elementary symmetric polynomial in $x_1,...,x_n$. So we know $s_1=a_1$. Then $s_2=\sum_{k\ne m} x_kx_m=(s_1^2-a_2)/2$. This gives $s_2$. Using $s_1^3$, $a_1,a_2, a_3$ allows us to compute $s_3$, etc.
Your condition shows that $y_1,...,y_n$ are roots of the same polynomial $f(x)$, so $(x_1,...,x_n)$ coincides with $(y_1,...,y_n)$ up to permutation.